Let $L$ and $M$ be two number fields. I know that $\renewcommand{\Gal}{\mathrm{Gal}}\Gal(LM/L)\cong \Gal(M/L\cap M)$ by restricting automorphisms. Fix a prime $p\in L\cap K$, and let $\frak{p_1}$ and $\frak{p_2}$ be primes in $L$ and $M$, respectively, both lying over $p$.
If $\frak{p_1}$ and $\frak{p_2}$ lie under $\frak{p}$ in $LM$, is it true that $ I(\frak{p}|\frak{p_1})\cong \text{$I$}(\frak{p_2}|\text{$p$}) $ and $ D(\frak{p}|\frak{p_1})\cong \text{$D$}(\frak{p_2}|\text{$p$})? $
I think that the only thing we need to check is that restriction does indeed map the above inertia/decomposition groups into one another. If this is true, then since we already know restriction gives an isomorphism on the ambient Galois groups, it must therefore descend to an isomorphism on the decomposition and inertia groups.
To check that inertia is mapped to inertia: if $\alpha\in \mathcal{O}_M$ and $\sigma\in I(\frak{p}|\frak{p_1})$ then $\sigma|_M\alpha-\alpha\in \frak{p}\cap\mathcal{O}_\text{$M$}=\frak{p_2}$ so $\sigma|_M\in \text{$I$}(\frak{p_2}|\text{$p$})$.
To check that decomposition is mapped to decomposition: if $\sigma\in D(\frak{p}|\frak{p_1})$ then $\sigma|_M \frak{p_2}=\sigma|_\text{$M$} \frak{p}\cap\mathcal{O}_\text{$M$}=\frak{p}\cap\mathcal{O}_\text{$M$}=\frak{p_2}$, so $\sigma|_M\in \text{$D$}(\frak{p_2}|\text{$p$})$.
Does this sound about right?